Given a 2D grid, each cell is either a wall 2, a zombie 1or people 0 (the number zero, one, two).Zombies can turn the nearest people(up/down/left/right) into zombies every day, but can not through wall. How long will it take to turn all people into zombies? Return -1 if can not turn all people into zombies.
Example
Given a matrix:
0 1 2 0 0
1 0 0 2 1
0 1 0 0 0
return 2
In this question, I store i * m + j in the queue, and m is the column number in the matrix.
In this question, pay attention that I should count the day first, becasue in the last day, everyone is zombie but if I don‘t return the day, the queue will keep iterate because it is not empty. For example, if I return the day number in the first day, the day should be 1. So add the day at first.
And pay attention that I can count the people first so if it drop to 0 I can return directly.
public class Solution {
/**
* @param grid a 2D integer grid
* @return an integer
*/
public int zombie(int[][] grid) {
if (grid == null || grid.length == 0 || grid[0].length == 0) {
return -1;
}
int n = grid.length;
int m = grid[0].length;
Queue<Integer> q = new LinkedList<>();
int people = 0;
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid[0].length; j++) {
if (grid[i][j] == 1) {
q.offer(i * m + j);
}
if (grid[i][j] == 0) {
people++;
}
}
}
int[] dx = {0, 0, 1, -1};
int[] dy = {1, -1, 0, 0};
int day = 0;
while (!q.isEmpty()) {
day++;
int size = q.size();
for (int i = 0; i < size; i++) {
int point = q.poll();
int x = point / m;
int y = point % m;
for (int j = 0; j < 4; j++) {
int nx = x + dx[j];
int ny = y + dy[j];
if (isValid(grid, nx, ny) && grid[nx][ny] == 0) {
q.offer(nx * m + ny);
grid[nx][ny] = 1;
people--;
if (people == 0) {
return day;
}
}
}
}
}
return -1;
}
private boolean isValid(int[][] grid, int x, int y) {
return x >= 0 && y >= 0 && x < grid.length && y < grid[0].length;
}
}